Sums of independent normal random variables, HELP!

Hey all!
So I'm studying for my final and this is the only problem in the chapter I cannot get and it's bugging the crap out of me
It is the section for sums of independent random variables, and follows the introduction of the following proof:

mean(X) = (1/n) (sum from i=1 to n of X), the sample mean of n independent N(u, o), is N(u, (o2/2)).
where u is the mean and o2 is the variance.

The problem itself is
Suppose the distribution of students' grades in a probability test is normal, with mean 72 and variance 25.
a) what is the probability that the average grade of such a class with 25 students is 75 or more?

b) if the prof teaches two different section of the course each with 25 students, what is the prob that the avg of one class is at least 3 more than the avg of the other class?

Now part a is straightforward, so i'm fine with that,
but part b involves a step which states that mean(X) and mean(Y) denote the means of the grades of the two classes, than mean(X) and mean(Y) and both N(72,1) which I did get from part a, but then my problem arises when it states that Since X and Y are both independent variables, and by the theorum listed above, than mean(X)-mean(Y) is N(0,2).
How in the world did they get N(0,2) from both of them being N(72,1) and independent, from the above theorum? I dont see the connection

also, N(0,2) or any two numbers denotes the normal variable notation that my book uses, if there was any confusion

I'm sorry if the post was not clear enough, as it was my first. ANY help would be appreciated

Re: Sums of independent normal random variables, HELP!

Hello,

First step : post in the correct subforum. The (advanced) statistics subforum suits better.
Second step : standardize your rv. That is to say you have to consider (X-u)/o, which follows a N(0,1). And then use a z-table to get the values you're looking for.